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In Duma's method 0.52g of an organic com...

In Duma's method 0.52g of an organic compound on combustion gave 68.6 mL `N_(2)` at `27^(@)C and 756mm` pressure. What is the percentage of nitrogen in the compound?

Text Solution

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`{:(V_1=48.6ml,,V_2=?),(P_1=756mm,,P_2=760mm):}`
`T_1=27+273=300K,T_2=273K`
Applying the general gas equation `(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`
we get The volume of nitrogen at STP,
`V_2=(P_1V_1)/(T_1)(T_2)/(T_1)`
`=(756xx48.6)/(300)xx(273)/(760)`
`=43.99ml`
Mass of organic compound `=0.45gm`
Precentage of nitrogen in the compound
`=(28)/(22400)xx(V_2)/(W)xx100`
`=(28)/(22400)xx(43.99)/(0.45)xx100`
`=12.22`
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